I'd like to find all unique binations of element positions of an array in JavaScript.

This is my array:

var places = ['x', 'y', 'z'];

I'd like to find the following binations: [0,1], [0,2], [1,2].

Currently I have the functional but slightly unwieldy:

for (var i = 0; i < places.length; i++) {
    for (var j = 0; j < places.length; j++) {
        if ((j > i) && (j != i)) { 
            console.log(i, j);
        }
    }
}

Is there a neater way to do it?

I'd like to find all unique binations of element positions of an array in JavaScript.

This is my array:

var places = ['x', 'y', 'z'];

I'd like to find the following binations: [0,1], [0,2], [1,2].

Currently I have the functional but slightly unwieldy:

for (var i = 0; i < places.length; i++) {
    for (var j = 0; j < places.length; j++) {
        if ((j > i) && (j != i)) { 
            console.log(i, j);
        }
    }
}

Is there a neater way to do it?

• javascript
• arrays
• permutation

• You don't need j!= i, every time you can reach it, it will evaluate false. any time j==i, j>i will evaluate to false, and the second condition will be skipped, along with rest of the 'if' statement. although, you could just use jodaka's suggestion and eliminate the if statement altogether – Chris Bye Commented Nov 3, 2011 at 13:50
• You're not paring the values of the array atm. If you want to pare the value, use places[i] and places[j] instead of i and j. – Rob W Commented Nov 3, 2011 at 13:51
• @RobW - sorry, I wasn't clear, by unique binations I mean unique binations of array positions, not of the values in the array. – Richard Commented Nov 4, 2011 at 18:09

2 Answers 2

// from codecademy.

var people = ["Alice", "Bob", "Carol", "Dave", "Ed"];
var n = people.length;
var i, j;

for(i = 0; i < n; i++){
    for(j = i + 1; j < n; j++){
        console.log(people[i] + ", " +  people[j]);
    }
}

// output
Alice, Bob
Alice, Carol
Alice, Dave
Alice, Ed
Bob, Carol
Bob, Dave
Bob, Ed
Carol, Dave
Carol, Ed
Dave, Ed

You can start j at i + 1 and eliminate your if condition.